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Popular >

extreme f(x)=sqrt(x^2-2x+2)

Solution

extreme points

f (x) = x^{2} - 2 x + 2

Solution

Minimum (1, 1)

Solution steps

f^{'} (x) = \frac{x - 1}{x ^{2} - 2 x + 2}

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 1

into

x^{2} - 2 x + 2 : 1

Minimum (1, 1)

Graph

Popular Examples

extreme f(x,y)=2x^2+2y^2-4x-y+1 extreme points

f (x, y) = 2 x^{2} + 2 y^{2} - 4 x - y + 1

extreme f(x,y)=sqrt(400-64x^2-49y^2)extreme points

f (x, y) = 400 - 64 x^{2} - 49 y^{2}

extreme f(x)=(x^2-5)^4(x^2+1)^5 extreme points

f (x) = (x^{2} - 5)^{4} (x^{2} + 1)^{5}

extreme f(x)=240x+57x^2-x^3 extreme points

f (x) = 240 x + 57 x^{2} - x^{3}

extreme f(x)=e^{-x^2-y^2}(x^2+2y^2)extreme points

f (x) = e^{- x^{2} - y^{2}} (x^{2} + 2 y^{2})

Frequently Asked Questions (FAQ)

What is the extreme f(x)=sqrt(x^2-2x+2) ?
The extreme f(x)=sqrt(x^2-2x+2) is Minimum(1,1)